In order to show a complete Matching of (U, V, E1), we will see that at the end of this game, the set of edges picked by Avoider satisfies Halls condition

Let (X,Y,E) be a bipartite graph

for every X U,

N(X) X

X has a perfect Matching.

XN(X) 19The trick A transpositionU V

Define a hypergraph

as following

1. the hyper vertices of

are the edges of E1.

2. The hyperedges of

are all the sets of edges between vertices of X U and Y V so that

XY n1.

XY 20The trick A transpositionU V

In the new graph, Enforcer

plays as Halls-Avoider,

and Avoider plays as Halls-Enforcer.

The new game (q, 1, )

Now - Halls-Avoider

applies the sufficient condition for winning the new game thus, not fully occupying any hyperedge of .

Later, we examine the size of the appropriate q

XY 21The trick A transpositionU V

Now well see that if Halls-Avoider doesnt fully occupy any hyperedge, Halls condition is satisfied by Halls-Enforcer.

At the end of the new game, take any subset X of U.

Look at N(X) which were picked by Halls-Enforcer.

This implies all the other edges (the new hyper vertices) were picked by Halls-Avoider.

N(X)XV-N(X) 22The trick A transpositionU V

As Halls-Avoider wins, it is impossible that

Xn-N(X) n1

(otherwise Halls-Avoider would have lost), and so

Xn-N(X) n

N(X) X

and so Hall condition is satisfied by Halls-Enforcer.

Now, retranspose

Halls-Enforcer is our Avoider

N(X)XV-N(X) 23About that q

Showing that Halls-Avoiders bias in the (q,1, ) game satisfies Avoiders sufficient condition at the start of the new game

24About that q 25About that q 26Remember The condition 27Perfect Matching over the edges of (1, q, H)U V

We saw for

Enforcer has a winning strategy!

We dont know whether this is a tight bound!

We dont know whether it is monotone!

Next, we play connectivity

28Connectivity on (1, q, H)

Where H contains q1 pairwise edge disjoint spanning trees(or more).

Avoider and Enforcer pick edges.

We will see that Enforcer can enforce Avoider to pick a spanning tree.

29Connectivity on (1, q, H)

Let T1, T2, , Tq1 be pairwise disjoint spanning trees of G(V,E).

Let W be the union (of the edges) of the q1 spanning trees.

Let L E\W.

30Enforcers strategy

Maintains in memory q1 acyclic graphs G1, G2, , Gq1.

In the beginning, Gi Ti (i1.q1) which are the original spanning trees.

31Enforcers strategy

Whenever Avoider picks an edge e of some Gj, Enforcer picks one edge fi of each Gi (i J), hence a total of q edges.

For all i j If Gi U e contains a (unique) cycle then fi is chosen as some unclaimed edge on this cycle. If Gi U e doesnt contains a cycle then fi is chosen arbitrarily.

In any case, Enforcer replaces Gi with Gi U e \ fi.

32Enforcers strategy

If Avoider picks an edge of L then Enforcer picks q edges of L. if there are not enough edges in L, Enforcer picks the rest of his acyclic graphs one edge of a different Gi.

If Enforcer starts the game, he would also start on the edges of L.

In any case, Enforcer removes his picks from the Gi.

33An example 34A winning strategy

Every unclaimed edge of W is in exactly one Gi, and every edge of W claimed by Avoider is in every Gi.

(an edge is added to Gi it was chosen by Avoider).

Every edge claimed by Enforcer is in no Gi.

(an edge is removed from Gi it was chosen by Enforcer).

35A winning strategy

After each round, each Gi is either a spanning tree, or a panning tree minus one edge.

It is obviously true at the beginning.

Suppose it is true after the Kth round.

If Avoider, on his K1 move, picks an edge e of some Gj, then Enforcer, on his K1 move, picks an edge fi of the other Gi.

If Gi U e is acyclic, then Gi U e \ fi is a spanning tree minus one edge.

If Gi U e contains a cycle then removing an edge fi from this cycle makes Gi U e \ fi a spanning tree over the same vertices as it was on the Kth round, thus it is a spanning tree or a spanning tree minus one edge.

36A winning strategy

In the end, each Gi has no unclaimed edges.

thus, all the edges of Gi are those which were picked by Avoider (as Enforcers picks are removed of each Gi and Avoiders picks are in each Gi).

So every Gi where is the set of edges picked by Avoider.

Gi in the end is a spanning tree or a spanning tree minus one edge.

We know that V - 1 because there are

(q1)(n-1) edges in W, so each player plays at least (n-1) rounds.

Thats why Gi must be a spanning tree, and Enforcer has won!

37Connectivity on (1, q, H)

We saw that if H contains q1 edge disjoint spreading trees, Enforcer has a winning strategy.

If H does not contain q1 edge disjoint spanning trees, and there is some limit on the number of total edges and the identity of the first player then, Avoider has a winning strategy.

There exists a polynomial time algorithm for finding q1 edge disjoint spanning trees, so Enforcers strategy is realistic.

Next we apply this outcome on special cases.

39Connectivity on complete graphs

On complete graphs, the game of (1, q, Kn ) has a threshold of

Kn contains edge disjoint spanning trees.

By what we have seen, Enforcer would win.

If q almost always Enforcer wins

Kn 40Connectivity on complete graphs

If q then each round total of are colored.

For Avoider to lose, the game must last at least n-1 rounds.

So total of edges.

If n is even then this is more than we have.

So Avoider wins.

Kn 41Connectivity on complete graphs

If n is odd and the graph has total edges.

In case Enforcer is the first player, we need

total edges.

which is still too much, so Avoider wins.

Kn 42Connectivity on complete graphs

If Avoider is the first player, we need

Which is exactly what we have.

Avoider wins only if he completes some cycle.

In an analogues game of Maker Breaker, it could be shown that Breaker wins.

So Avoider loses in this case.

Kn 43Connectivity on complete graphs

We saw the threshold is

except for special cases where n is odd and Avoider starts, and then the threshold is

Another way of putting it is that this game is monotone.

Kn 44Connectivity on complete graphs

Another special game is the game of (1,1 Kn).

Enforcer can enforce Avoider to build spanning trees.

The strategy playing

separate games, each on a graph with 2 edge disjoint spanning trees.

Kn 45Summery

Avoider Enforcer games are not generally monotone.

A sufficient condition for Avoider to win in (p,q,H), which was not tight.

We studied 2 special games

Perfect Matching in (1,q,Kn,n), where Enforcer has a winning strategy for q of order n/log2n. We dont know whether there exists a threshold, and if so, of what ordeer.

46Summery

Connectivity we saw a general sufficient condition for Enforcer to enforce a spanning tree in (1,q,H), where H contains q1 edge disjoint spanning trees.

Connectivity in (1,q,Kn), which is a special case of a monotone game with a threshold of

Connectivity in (1,1,Kn), in which Enforcer can enforce Avoider to build edge disjoint spanning trees.

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